Finding Formulas from Systematic Names: How Does the Number in Parentheses Help?

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SUMMARY

The discussion focuses on determining chemical formulas from systematic names, specifically using oxidation numbers indicated in parentheses. For example, sodium chlorate(V) corresponds to NaClO3, while sodium chlorate(I) results in NaClO. The oxidation states of chlorine in these compounds are +5 and +1, respectively. Participants clarify that the nomenclature system differs from traditional "ate" and "ite" classifications, emphasizing the importance of understanding oxidation states for accurate formula derivation.

PREREQUISITES
  • Understanding of oxidation states in chemistry
  • Familiarity with systematic naming conventions for chemical compounds
  • Basic knowledge of ionic compounds and their charges
  • Proficiency in balancing chemical equations
NEXT STEPS
  • Research the systematic naming conventions for polyatomic ions
  • Learn about oxidation states and their implications in chemical formulas
  • Study the differences between "ate," "ite," and other related nomenclature systems
  • Explore examples of calculating oxidation states in various chemical compounds
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Chemistry students, educators, and professionals involved in chemical nomenclature and formula derivation will benefit from this discussion.

IDK10
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Homework Statement


Sodium chlorate(I)
Sodium chlorate(V)
Potassium nitride(III)
Phosphorus(III) chloride
Magnesium iodate(I)

Homework Equations


None

The Attempt at a Solution


I know the number in the bracket refers to what's before it, i.e. in sodium chlorate(V), the (V) means that the chlorine has an oxidation number of +5, but how does this help me find the fomula of the copound.

For example with sodium chlorate(V):
I know its made from sodium (Na+) ions, and chlorate(ClO3-) ions.
 
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Molecule must be neutral, assume charge on oxygen to be -2.
 
I'm not familiar with the use of this style for negative oxidation numbers, e.g. potassium nitride(III), if this is really K3N and not a typo for potassium nitrate(III) = potassium nitrite = KNO2.
 
Borek said:
Molecule must be neutral, assume charge on oxygen to be -2.
For sodium chlorate(I)
Na would be +1
Cl would be +1
and O would be -2
1 + 1 - 3(2) = -4 (what would I do with this?)

Or

Na + 1 -6 = 0
Na = 5 (?)

For sodium chlorate(V)
Na would be +1
Cl would be +5
and O would be -2
1 + 5 - 3(-2) = 0
NaClO3
 
.
 
How many oxygens are there in sodium chlorate(I)?
Note: in this nomenclature you must not get hung up on the notion that e.g. "chlorate" means "ClO3". All oxysalts are "ates" with different oxidation number of the central atom. This is quite different from the "ate", "ite", "per...ate" system.
 
mjc123 said:
How many oxygens are there in sodium chlorate(I)?
Note: in this nomenclature you must not get hung up on the notion that e.g. "chlorate" means "ClO3". All oxysalts are "ates" with different oxidation number of the central atom. This is quite different from the "ate", "ite", "per...ate" system.
I'm not sure, I have to find the formula just from the name. I know sodium chlorate(v) has 3.
 
IDK10 said:
For sodium chlorate(I)
Na would be +1
Cl would be +1
and O would be -2

And assuming one Na and one Cl, how many oxygens are needed for the molecule to be neutral?
 
Borek said:
And assuming one Na and one Cl, how many oxygens are needed for the molecule to be neutral?
While I was away, amd therefore didn't see your reply, I was thinking and I think i got it:
1 + 1 -2O = 0
-2O = -2
O = 1
NaClO?
 
  • #10
IDK10 said:
While I was away, amd therefore didn't see your reply, I was thinking and I think i got it:
1 + 1 -2O = 0
-2O = -2
O = 1
NaClO?

Yep.

Try with chlorate(III) and chlorate(V).
 
  • #11
Borek said:
Yep.

Try with chlorate(III) and chlorate(V).
Sodium chlorate(III):
1+3-2O=0
-2O=-4
O=2
NaClO2

Sodium chlorate(V):
1+5-2O=0
-2O=-6
O=3
NaClO3
 
  • #12
At least with these you should have no problems now :wink:
 

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